Question

g student pushes a 80 Kg crate across the classroom floor a distance 5 m at constant speed of 0.4 m/s . If the coefficient of kinetic friction between the crate and the floor is 0.9, how much work did she do

Answer 1

Answer: The work done is W = 3528J.

Explanation: Work (W) is an amount of energy used to do determined movement. To calculate it, we use the formula, as the movement is horizontal:

W = F.d, where

F is force on the object

d is distance it went through.

The force acting on the crate is due to friction.

$F_{f}$ = μ . $F_{N}$

μ is the kinetic friction

$F_{N}$ is the normal force

There is no up and down movement, so $F_{N} = F_{g}$, with g = 9.8m/s²

$F_{f}$ = μ . m . g

$F_{f}$ = 0.9 . 80 . 9.8

$F_{f}$ = 705.6 N

Substituing $F_{f}$  into the work formula, we have:

W = $F_{f}$ . d

W = 705.6 . 5

W = 3528 J

The work done to move the crate across a distance of 5m is 3528 J.